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Ta có \(a^2+\dfrac{1}{b+c}=a^2+\dfrac{1}{6-a}\)
Mà \(a+b+c=6\Rightarrow0\le a,b,c\le2\)
\(\Rightarrow a^2+\dfrac{1}{6-a}\ge2^2+\dfrac{1}{6-2}=\dfrac{17}{4}\)
\(\Rightarrow P=\sum\sqrt{a^2+\dfrac{1}{b+c}}=\sum\sqrt{a^2+\dfrac{1}{6-a}}\ge\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}=\dfrac{3\sqrt{17}}{2}\)
Dấu \("="\Leftrightarrow a=b=c=2\)
Áp dụng BĐT Bunhiacopxki, có:
\(\hept{\begin{cases}\sqrt{a^2+\frac{1}{a^2}}=\frac{1}{\sqrt{17}}\sqrt{\left(a^2+\frac{1}{a^2}\right)\left(4^2+1^2\right)}\ge\frac{1}{\sqrt{17}}\left(4a+\frac{1}{a}\right)\\\sqrt{b^2+\frac{1}{c^2}}=\frac{1}{\sqrt{17}}\sqrt{\left(b^2+\frac{1}{b^2}\right)\left(4^2+1^2\right)}\ge\frac{1}{\sqrt{17}}\left(4b+\frac{1}{b}\right)\end{cases}}\)
Lúc này được \(P\ge\frac{1}{\sqrt{17}}[4\left(a+b\right)+\left(\frac{1}{a}+\frac{1}{b}\right)]\)
Thấy \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
Áp dụng BĐT Cauchy và đề bài được
\(P\ge\frac{1}{\sqrt{17}}[4\left(a+b\right)+\frac{4}{a+b}]\)
\(=\frac{1}{\sqrt{17}}[\frac{a+b}{4}+\frac{4}{a+b}+\frac{15\left(a+b\right)}{4}]\ge\frac{1}{\sqrt{17}}[2+15]=\sqrt{17}\)
Dấu "=" xảy ra khi:
\(\hept{\begin{cases}\frac{a}{4}=\frac{1}{a}\\\frac{b}{4}=\frac{1}{b}\end{cases}}\)
\(\Leftrightarrow a=b=2\)
Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=1\)
BĐT trở thành: \(\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}+\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}+\dfrac{zx}{\sqrt{x^2+z^2+2y^2}}\le\dfrac{1}{2}\)
Ta có:
\(x^2+z^2+y^2+z^2\ge\dfrac{1}{2}\left(x+z\right)^2+\dfrac{1}{2}\left(y+z\right)^2\ge\left(x+z\right)\left(y+z\right)\)
\(\Rightarrow\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}\le\dfrac{xy}{\sqrt{\left(x+z\right)\left(y+z\right)}}\le\dfrac{1}{2}\left(\dfrac{xy}{x+z}+\dfrac{xy}{y+z}\right)\)
Tương tự: \(\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}\le\dfrac{1}{2}\left(\dfrac{yz}{x+y}+\dfrac{yz}{x+z}\right)\)
\(\dfrac{zx}{\sqrt{z^2+x^2+2y^2}}\le\dfrac{1}{2}\left(\dfrac{zx}{x+y}+\dfrac{zx}{y+z}\right)\)
Cộng vế với vế:
\(VT\le\dfrac{1}{2}\left(\dfrac{zx+yz}{x+y}+\dfrac{xy+zx}{y+z}+\dfrac{yz+xy}{z+x}\right)=\dfrac{1}{2}\left(x+y+z\right)=\dfrac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c\)
Dễ dàng c/m : \(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}=1\)
Ta có : \(\dfrac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le\dfrac{1}{a+b+4}\le\dfrac{1}{4}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}\right)\)
Suy ra : \(\Sigma\dfrac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le2.\dfrac{1}{4}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\right)=\dfrac{1}{2}.1=\dfrac{1}{2}\)
" = " \(\Leftrightarrow a=b=c=1\)
Do 1/b+1/c=3/4-1/a suy ra \(\sum\) (1a/)=3/4
Ta có \(\dfrac{\sqrt{b^2+bc+c^2}}{a^2}\)= \(\dfrac{\sqrt{\left(b+c\right)^2-bc}}{a^2}\ge\dfrac{\sqrt{\left(b+c\right)^2-\dfrac{\left(b+c\right)^2}{4}}}{a^2}=\dfrac{\sqrt{3}\left(b+c\right)}{2a^2}\)
Tương tự ta được:
P\(\ge\) \(\sqrt{3}\) \(\left(\sum\dfrac{b+c}{a^2}\right)\) \(\ge\) \(\sqrt{3}\) (1/a+1/b+1/c) \(\ge\dfrac{3\sqrt{3}}{4}\)
Đẳng thức xảy ra \(\Leftrightarrow\) a=b=c=4
Với mọi \(0< a< \dfrac{1}{2}\) ta có:
\(\left(\sqrt{2a}-1\right)^2\ge0\Rightarrow2a+1\ge2\sqrt{2a}\)
\(\Rightarrow1\ge2\sqrt{a}\left(\sqrt{2}-\sqrt{a}\right)\)
\(\Rightarrow\dfrac{1}{\sqrt{2}-\sqrt{a}}\ge2\sqrt{a}\)
Do đó:
\(\dfrac{2+\sqrt{2a}}{2-a}=\dfrac{2-a+a+\sqrt{2a}}{2-a}=1+\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{2}\right)}{\left(\sqrt{2}-\sqrt{a}\right)\left(\sqrt{2}+\sqrt{a}\right)}=1+\dfrac{\sqrt{a}}{\sqrt{2}-\sqrt{a}}\ge1+\sqrt{a}.2\sqrt{a}=2a+1\)
Tương tự:
\(\dfrac{2+\sqrt{2b}}{2-b}\ge2b+1\)
Cộng vế:
\(\dfrac{2+\sqrt{2a}}{2-a}+\dfrac{2+\sqrt{2b}}{2-b}\ge2a+1+2b+1=4\) (đpcm)
Dấu "=" xảy ra khi \(a=b=\dfrac{1}{2}\)
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\((a^2+\frac{1}{b^2})(1+4^2)\geq (a+\frac{4}{b})^2\Rightarrow \sqrt{a^2+\frac{1}{b^2}}\geq \frac{1}{\sqrt{17}}(a+\frac{4}{b})\)
Hoàn toàn tương tự với những cái còn lại và cộng theo vế suy ra:
$S\geq \frac{1}{\sqrt{17}}(a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c})$
$\geq \frac{1}{\sqrt{17}}(a+b+c+\frac{36}{a+b+c})$ theo BĐT Cauchy-Schwarz.
Áp dụng BĐT AM-GM:
\(a+b+c+\frac{9}{4(a+b+c)}\geq 3\)
\(\frac{135}{4(a+b+c)}\geq \frac{135}{4.\frac{3}{2}}=\frac{45}{2}\)
\(\Rightarrow a+b+c+\frac{36}{a+b+c}\geq \frac{51}{2}\)
\(\Rightarrow S\geq \frac{3\sqrt{17}}{2}\)
Vậy $S_{\min}=\frac{3\sqrt{17}}{2}$
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\((a^2+\frac{1}{b^2})(1+4^2)\geq (a+\frac{4}{b})^2\Rightarrow \sqrt{a^2+\frac{1}{b^2}}\geq \frac{1}{\sqrt{17}}(a+\frac{4}{b})\)
Hoàn toàn tương tự với những cái còn lại và cộng theo vế suy ra:
$S\geq \frac{1}{\sqrt{17}}(a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c})$
$\geq \frac{1}{\sqrt{17}}(a+b+c+\frac{36}{a+b+c})$ theo BĐT Cauchy-Schwarz.
Áp dụng BĐT AM-GM:
\(a+b+c+\frac{9}{4(a+b+c)}\geq 3\)
\(\frac{135}{4(a+b+c)}\geq \frac{135}{4.\frac{3}{2}}=\frac{45}{2}\)
\(\Rightarrow a+b+c+\frac{36}{a+b+c}\geq \frac{51}{2}\)
\(\Rightarrow S\geq \frac{3\sqrt{17}}{2}\)
Vậy $S_{\min}=\frac{3\sqrt{17}}{2}$
Bài 1:
Áp dụng BĐT Bunhiacopxky ta có:
$(a^2+b^2+c^2)(1+1+1)\geq (a+b+c)^2$
$\Leftrightarrow 3(a^2+b^2+c^2)\geq 1$
$\Leftrightarrow a^2+b^2+c^2\geq \frac{1}{3}$ (đpcm)
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
Bài 2:
Áp dụng BĐT Bunhiacopxky:
$(a^2+4b^2+9c^2)(1+\frac{1}{4}+\frac{1}{9})\geq (a+b+c)^2$
$\Leftrightarrow 2015.\frac{49}{36}\geq (a+b+c)^2$
$\Leftrightarrow \frac{98735}{36}\geq (a+b+c)^2$
$\Rightarrow a+b+c\leq \frac{7\sqrt{2015}}{6}$ chứ không phải $\frac{\sqrt{14}}{6}$ :''>>
Áp dụng BĐT Minicopski, ta có:
\(P=\sqrt{a^2+\dfrac{1}{a^2}}+\sqrt{b^2+\dfrac{1}{b^2}}\ge\sqrt{\left(a+b\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2}\\ \Rightarrow P\ge\sqrt{4^2+\left(\dfrac{4}{a+b}\right)^2}=\sqrt{16+\left(\dfrac{4}{4}\right)^2}=\sqrt{17}\)
Đẳng thức xảy ra \(\Leftrightarrow a=b=2\)
Áp dụng BĐT Cô si
⇒ P≥ \(\sqrt{2\sqrt{a^2.\dfrac{1}{a^2}}}+\sqrt{2\sqrt{b^2.\dfrac{1}{b^2}}}\)
\(=\sqrt{2}+\sqrt{2}\)
\(=2\sqrt{2}\)